To evaluate the integral: \(\int_1^2\int_0^1(3x^2+4y^3)dydx\)

Solution:

When we integrate with respect to one variable then other will be kept as constant.

Evaluating the integral.

\(\int_1^2\int_0^1(3x^2+4y^3)dy=\int_1^2[\int_0^1(3x^2+4y^3)dy]dx\)

\(=\int_1^2 [(3x^2y+4\cdot\frac{y^4}{4})_0^1]dx\)

\(=\int_1^2 [3x^2\cdot 1+1^4-0]dx\)

\(=\int_1^2 (3x^2+1)dx\)

\(=[3\cdot\frac{x^3}{3}+x]_1^2\)

\(=[x^3+x]_1^2\)

\(=[(2^3+2)-(1^3+1)]\)

\(=[12-2]\)

\(=10\)

\(\text{Hence, required answer is 10}\)

Solution:

When we integrate with respect to one variable then other will be kept as constant.

Evaluating the integral.

\(\int_1^2\int_0^1(3x^2+4y^3)dy=\int_1^2[\int_0^1(3x^2+4y^3)dy]dx\)

\(=\int_1^2 [(3x^2y+4\cdot\frac{y^4}{4})_0^1]dx\)

\(=\int_1^2 [3x^2\cdot 1+1^4-0]dx\)

\(=\int_1^2 (3x^2+1)dx\)

\(=[3\cdot\frac{x^3}{3}+x]_1^2\)

\(=[x^3+x]_1^2\)

\(=[(2^3+2)-(1^3+1)]\)

\(=[12-2]\)

\(=10\)

\(\text{Hence, required answer is 10}\)